document.write( "Question 76462: 3r-4s=0 and 2r+5s=23 \n" ); document.write( "
Algebra.Com's Answer #54882 by checkley75(3666)\"\" \"About 
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3R-4S=0
\n" ); document.write( "3R=4S
\n" ); document.write( "R=4S/3 NOW SUBSTITUTE (4S/3) FOR R ANS SOLVE FOR S
\n" ); document.write( "2(4S/3)+5S=23
\n" ); document.write( "12S/3+5S=23
\n" ); document.write( "(12S+3*5S)/3=23 NOW CROSS MULTIPLY
\n" ); document.write( "12S+15S=3*23
\n" ); document.write( "27S=69
\n" ); document.write( "S=69/27
\n" ); document.write( "S=23/9 ANSWER.
\n" ); document.write( "R=(4*23/9)/3
\n" ); document.write( "R=92/9*1/3
\n" ); document.write( "R=92/27\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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